Problem: Consider the polar curve $r=2+6\sin(\theta)$. What is the slope of the tangent line to the curve $r$ when $\theta = \dfrac{\pi}{6}$ ? Give an exact expression. $\text{slope }=$
The slope of the tangent line at a point is equal to $\dfrac{dy}{dx}$ at that point. In the case of polar curves, we can use the relationship: $\dfrac{dy}{dx}=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)}$ For a polar curve, $x={r}\cos(\theta)$ and $y={r}\sin(\theta)$. Therefore, in our problem we have: $\begin{aligned} x&={(2+6\sin(\theta))}\cos(\theta) \\\\ y&={(2+6\sin(\theta))} \sin(\theta) \end{aligned}$ Let's find $\dfrac{dy}{dx}$. $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)} \\\\ &=\dfrac{6\sin(2\theta)+2\cos(\theta)}{6\cos(2\theta)-2\sin(\theta)} \\\\ &=\dfrac{3\sin(2\theta)+\cos(\theta)}{3\cos(2\theta)-\sin(\theta)} \end{aligned}$ Finally, we evaluate $\dfrac{dy}{dx}$ at ${\theta = \dfrac{\pi}{6}}$. $\begin{aligned} \left. \dfrac{dy}{dx}\right| _{\theta =\tfrac{\pi }{6}}&=\dfrac{3\sin\left(2\left({\dfrac{\pi}{6}}\right)\right)+\cos\left({\dfrac{\pi}{6}}\right)}{3\cos\left(2\left({\dfrac{\pi}{6}}\right)\right)-\sin\left({\dfrac{\pi}{6}}\right)} \\\\ &=\dfrac{3\sin\left(\dfrac{\pi}{3}\right)+\cos\left(\dfrac{\pi}{6}\right)}{3\cos\left(\dfrac{\pi}{3}\right)-\sin\left(\dfrac{\pi}{6}\right)} \\\\ &=\dfrac{3\left(\dfrac{\sqrt{3}}{2}\right)+\left(\dfrac{\sqrt{3}}{2}\right)}{3\left(\dfrac{1}{2}\right)-\left(\dfrac{1}{2}\right)} \\\\ &=\dfrac{4\sqrt{3}}{2} \\\\ &=2\sqrt{3} \end{aligned}$ The slope of the tangent line to the curve $r$ when $\theta=\dfrac{\pi}{6}$ equals $2\sqrt{3}$. The graph of the tangent is shown.